Re: Need soln explaination

[rshotwel@ivy.tec.in.us (Allen Shotwell)]

I'm sure I'm missing something here, but here it goes.

Before the masses are released,
Energy = potential energy of masses+ kinetic energy of masses = 3gh+2gh + 0


After the masses are released (and 3kg falls .5 m),
Energy = potential energy of masses + kinetic energy of masses = 3g(h-.5) +
2g(h+.5) + (1/2)3v^2 + (1/2)2v^2

Energy before = energy after

3gh+2gh+0 = 3g(h-.5) + 2g(h+.5) + (1/2)3v^2 + (1/2)2v^2

5gh = 5gh - .5g + 2.5v^2

-.5g + 2.5v^2 = 0

v = 1.98 m/s

All this makes the usual introductory text assumptions about friction,
energy loss, etc.

> I got a problem out of an older physics text (one without a solutions
> manual). My students and spent a while debating the method of attaining
> the solution. Can someone explain how they would solve it -using energy.
>
> PROBLEM
> Masses of 3 kg and 2 kg are hanging at opposite ends of a cord that
> passes over a frictionless pulley. The system is held stationary for a
> while, then released. What will be the speed of the masses when the 3 kg
> mass has descended 0.5 m below its starting point? (USE ENERGY
> RELATIONSHIPS TO SOLVE.)
>
> -tony
R. Allen Shotwell
Chair, Science and Math
Ivy Tech State College
Terre Haute, IN USA